Thales Theorem - Celestial Tutors

This is a special case of inscribed angle theorem, an important theorem of circle geometry. This theorem can be stated in any of the following ways:

If three points A, B, and C lie on the circumference of a circle, where the line AC is the diameter of the circle, then the angle ∠ABC is a right angle (90°)

The diameter of a circle always subtends a right angle to any point on the circle.

An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle.

Here is an interesting and very easy proof of Thales theorem:

To prove this we use following two facts:

– Sum of all angles of a triangle is 180.

– Base angles of an isosceles triangle are equal.

In this circle we have :

$\overline{OA}=\overline{OB}=\overline{OC}$ reason being all are radii of same circle.

$\angle OAB=\angle OBA =\alpha$ reason being base angles of isosceles triangle OAB

Similarly ,

$\angle OBC=\angle OCB =\beta$ reason being base angles of isosceles triangle OBC

Also, $\angle ABC=\alpha +\beta$

Since sum of all angles of a triangle is 180 , so

$\angle ABC+\angle BCA +\angle CAB = 180$

$(\alpha +\beta )+\beta +\alpha = 180$

$2(\alpha +\beta )= 180$

$(\alpha +\beta )= 90$

$\therefore \angle ABC= 90^{\circ}$

Here are some examples on Thales Theorem:

1. In the given circle FH is a diameter passing through its center J. Find x and angle HFG.

Since FH is a diameter so using Thales theorem, FGH would be a right triangle with right angle at point G.

Therefore we can apply triangle’s angle sum property,

$\angle HFG +\angle FGH +\angle FHG = 180$

$4x+2 +90 +9x-3 = 180$

$4x+9x+2-3+90 = 180$

$13x+89 = 180$

$13x = 91$

$x = 7$

$\therefore \angle HFG = 4x+2 = 4(7)+2= 30$

$\angle HFG = 30^{\circ}$

2. In the given circle, find the measure of angle PQR . Assume point R is the center of the circle.

Since R is given center of circle, PS would be the diameter of the circle.

Therefore applying Thales theorem for triangle PQS, angle PQS would be a right angle.

$\angle PQS=90$

Since RS = RQ (radii of same circle) , base angles of isosceles triangle RQS would be equal.

$\angle RQS=\angle RSQ = 24$

Also

$\angle PQS = \angle PQR +\angle RQS$

$90 = \angle PQR + 24$

$66 = \angle PQR$

Hence the measure of angle PQR = 66

3. In the circle shown, BC is a diameter with center A. Given m∠ABE = 26 and m∠ADB=18

Find a) m∠DAB b) m∠BAE c) m∠DAE

Since A is the center of circle, so AB=AD=AE , as all are radii of same circle.

Therefore ABD is an isosceles triangle and hence base angles would be equal.

$\angle ADB=\angle ABD = 18$

Using triangle’s angle sum property,

$\angle ABD +\angle ADB +\angle DAB = 180$

$18+18 +\angle DAB = 180$

$\angle DAB = 180-(18+18)$

$\angle DAB = 144^{\circ}$

b) Using exactly same concept for triangle ABE, we have

$\angle AEB=\angle ABE = 26$

Using triangle’s angle sum property,

$\angle ABE +\angle AEB +\angle BAE = 180$

$26+26 +\angle BAE = 180$

$\angle BAE = 180-(26+26)$

$\angle BAE = 128^{\circ}$

c) We see that three angles make a complete angle of 360 at point A.

$\angle DAB +\angle BAE +\angle DAE = 360$

$144 +128 +\angle DAE = 360$

$\angle DAE = 360-(144 +128)$

$\angle DAE = 88^{\circ}$

4. In the figure below, O is the center of circle and AD is a diameter.

a. Find 𝑚∠𝐴OB.
b. If 𝑚∠BOC ∶ 𝑚∠𝐶OD = 6:5 , what is 𝑚∠𝐵OC?

Since OB = OD being radii of same circle,

$\therefore \angle OBD = \angle ODB$ as base angles of isoscele triangles are equal.

$\therefore \angle OBD = \angle ODB=24$

Using exterior angle property of triangles, exterior angle is equal to sum of opposite interior angles.

$\therefore ext\angle AOB = 24+24= 48$

$\angle BOD = 180-48= 132$

$\angle BOD= \angle BOC+\angle COD$

These two angles are given in the ratio 6 : 5

$\therefore 6x+5x= 132$

11x= 132

x= 12

$\angle BOC = 6x = 6*12 = 72$

$\angle BOC = 72^{\circ}$

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